NCERT Solutions for Class 11 Chemistry Chapter 2
Structure of Atom Class 11
Chapter 2 Structure of Atom Exercise Solutions
Exercise : Solutions of Questions on Page Number : 65
(i) Calculate the number of electrons which will together weigh one gram.
Answer :
(i) Mass of one electron = 9.10939 ื 10โ31 kg
Number of electrons that weigh 9.10939 ื 10โ31 kg = 1 Number of electrons that will weigh 1 g = (1 ื 10โ3 kg)
= 0.1098 ื 10โ3 + 31
= 0.1098 ื 1028
= 1.098 ื 1027
(ii) Mass of one electron = 9.10939 ื 10โ31 kg
Mass of one mole of electron = (6.022 ื 1023) ื (9.10939 ื10โ31 kg)
= 5.48 ื 10โ7 kg
Charge on one electron = 1.6022 ื 10โ19 coulomb
Charge on one mole of electron = (1.6022 ื 10โ19 C) (6.022 ื 1023)
(i) Calculate the total number of electrons present in one mole of methane.
(iii) Find (a) the total number and (b) the total mass of protons in 34 mg of NH3at STP. Will the answer change if the temperature and pressure are changed?
(i) Number of electrons present in 1 molecule of methane (CH4)
{1(6) + 4(1)} = 10
Number of electrons present in 1 mole i.e., 6.023 ื 1023 molecules of methane
= 6.022 ื 1023 ื 10 = 6.022 ื 1024
(ii) (a) Number of atoms of 14C in 1 mole= 6.023 ื 1023
Since 1 atom of 14C contains (14 โ 6) i.e., 8 neutrons, the number of neutrons in 14 g of 14C is (6.023 ื 1023) ื8. Or, 14 g of 14C contains (6.022 ื 1023 ื 8) neutrons.
= 2.4092 ื 1021
(b) Mass of one neutron = 1.67493 ื 10โ27 kg Mass of total neutrons in 7 g of 14C
= (2.4092 ื 1021) (1.67493 ื 10โ27 kg)
= 4.0352 ื 10โ6 kg
(iii) (a) 1 mole of NH3 = {1(14) + 3(1)} g of NH3
= 17 g of NH3
= 6.022ื 1023 molecules of NH3
Total number of protons present in 1 molecule of NH3
= {1(7) + 3(1)}
= 10
Number of protons in 6.023 ื 1023 molecules of NH3
= (6.023 ื 1023) (10)
= 6.023 ื 1024
⇒ 17 g of NH3 contains (6.023 ื 1024) protons. Number of protons in 34 mg of NH3
= 1.2046 ื 1022
(b) Mass of one proton = 1.67493 ื 10โ27 kg Total mass of protons in 34 mg of NH3
= (1.67493 ื 10โ27 kg) (1.2046 ื 1022)
= 2.0176 ื 10โ5 kg
The number of protons, electrons, and neutrons in an atom is independent of temperature and pressure conditions. Hence, the obtained values will remain unchanged if the temperature and pressure is changed.
How many neutrons and protons are there in the following nuclei? , , , ,
6
13 C:
Atomic mass = 13
Atomic number = Number of protons = 6
Number of neutrons = (Atomic mass) โ (Atomic number)
= 13 โ 6 = 7
:
Atomic mass = 16 Atomic number = 8 Number of protons = 8
Number of neutrons = (Atomic mass) โ (Atomic number)
= 16 โ 8 = 8 :
Atomic mass = 24
Atomic number = Number of protons = 12
Number of neutrons = (Atomic mass) โ (Atomic number)
= 24 โ 12 = 12
:
Atomic mass = 56
Atomic number = Number of protons = 26
Number of neutrons = (Atomic mass) โ (Atomic number)
= 56 โ 26 = 30
:
Atomic mass = 88
Atomic number = Number of protons = 38
Number of neutrons = (Atomic mass) โ (Atomic number)
Write the complete symbol for the atom with the given atomic number (Z) andAtomic mass (A) (i) Z = 17, A = 35
(iii) Z = 4, A = 9
Yellow light emitted from a sodium lamp has a wavelength (รยป) of 580 nm. Calculate the frequency (รยฝ) and wave number ( ) of the yellow light.
From the expression,
We get,
.. (i)
Where,
รยฝ = frequency of yellow light
c = velocity of light in vacuum = 3 ื 108 m/s
รยป = wavelength of yellow light = 580 nm = 580 ื 10โ9m Substituting the values in expression (i):
Thus, frequency of yellow light emitted from the sodium lamp
= 5.17 ื 1014 sโ1
Wave number of yellow light,
Find energy of each of the photons which
(ii) have wavelength of 0.50 รฆ.
(i) Energy (E) of a photon is given by the expression,
E =
Where,
h = Planck's constant = 6.626 ื 10โ34Js
รยฝ = frequency of light = 3 ื 1015Hz Substituting the values in the given expression ofE:
E = (6.626 ื 10โ34) (3 ื 1015)
E = 1.988 ื 10โ18J
(ii) Energy (E) of a photon having wavelength (รยป) is given by the expression,
h = Planck's constant = 6.626 ื 10โ34Js
c = velocity of light in vacuum = 3 ื 108m/s Substituting the values in the given expression of E:
Calculate the wavelength, frequency and wave number of a light wave whose period is 2.0 x 10-10s.
Frequency (รยฝ) of light
Wavelength (รยป) of light Where,
c = velocity of light in vacuum = 3ื108m/s
Substituting the value in the given expression of รยป:
What is the number of photons of light with a wavelength of 4000 pm that provide 1 J of energy?
Energy (E) of a photon = hรยฝ Energy (En) of 'n' photons = nhรยฝ
Where,
รยป = wavelength of light = 4000 pm = 4000 ื10โ12m c = velocity of light in vacuum = 3 ื 108 m/s
h = Planck's constant = 6.626 ื 10โ34Js Substituting the values in the given expression of n:
A photon of wavelength 4 x 10-7m strikes on metal surface, the work function of the metal being 2.13 eV. Calculate (i) the energy of the photon (eV), (ii) the kinetic energy of the emission, and (iii) the velocity of the photoelectron (1 eV= 1.6020 x 10-19J).
(i) Energy (E) of a photon = hรยฝ Where,
h = Planck's constant = 6.626 ื 10โ34Js
c = velocity of light in vacuum = 3 ื 108 m/s รยป = wavelength of photon = 4 ื 10โ7m
Substituting the values in the given expression of E:
Hence, the energy of the photon is 4.97 ื 10โ19J.
(ii) The kinetic energy of emission Ekis given by
= (3.1020 โ 2.13) eV
= 0.9720 eV
Hence, the kinetic energy of emission is 0.97 eV.
(iii) The velocity of a photoelectron (รยฝ) can be calculated by the expression,
v = 5.84 ื 105 msโ1
Hence, the velocity of the photoelectron is 5.84 ื 105msโ1.
Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in kJ mol-1.
Energy of sodium (E)
= 4.947 ื 105 J molโ1
= 494.7 ื 103 J molโ1
A 25 watt bulb emits monochromatic yellow light of wavelength of 0.57รยผm. Calculate the rate of emission of quanta per second.
Power of bulb, P= 25 Watt = 25 Jsโ1
Energy of one photon, E= hรยฝ
Substituting the values in the given expression of E:
E = 34.87 ื 10โ20 J
Rate of emission of quanta per second
Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 รโฆ. Calculate threshold frequency ( ) and work function (W0) of the metal.
Threshold wavelength of radian = 6800 ื 10โ10m Threshold frequency of the metal
= 4.41 ื 1014sโ1
Thus, the threshold frequency of the metal is 4.41 ื 1014sโ1. Hence, work function (W0) of the metal = hรยฝ0
= (6.626 ื 10โ34Js) (4.41 ื 1014sโ1)
What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with n = 4 to an energy level with n = 2?
Substituting the values in the given expression of E:
E = โ (4.0875 ื 10โ19 J)
The negative sign indicates the energy of emission.
Wavelength of light emitted
Substituting the values in the given expression of รยป:
How much energy is required to ionise a H atom if the electron occupies n = 5 orbit? Compare your answer with the ionization enthalpy of H atom (energy required to remove the electron from n =1 orbit).
The expression of energy is given by,
Where,
Z = atomic number of the atom
n= principal quantum number
For ionization from n1= 5 to ,
Hence, the energy required for ionization from n= 5 to n = is 8.72 ื 10โ20J. Energy required for n1= 1 to n = ,
What is the maximum number of emission lines when the excited electron of an H atom in n = 6 drops to the ground state?
Hence, a total number of (5 + 4 + 3 + 2 + 1) 15 lines will be obtained in the emission spectrum.
The number of spectral lines produced when an electron in the nthlevel drops down to the ground state is given by
.
Given,
n= 6
Number of spectral lines = 15
(i) The energy associated with the first orbit in the hydrogen atom is -2.18 x 10-18J atom-1. What is the energy associated with the fifth orbit?
Answer :
(i) Energy associated with the fifth orbit of hydrogen atom is calculated as:
E5= โ8.72 ื 10โ20J
(ii) Radius of Bohr's nthorbit for hydrogen atom is given by,
rn= (0.0529 nm) n2
For,
n = 5
r5= (0.0529 nm) (5)2
Calculate the wave number for the longest wavelength transition in the Balmer series of atomic hydrogen.
For the Balmer series, ni = 2. Thus, the expression of wavenumber is given by,
Wave number is inversely proportional to wavelength of transition. Hence, for the longest wavelength transition, has to be the smallest.
For to be minimum, nf should be minimum. For the Balmer series, a transition from ni = 2 to nf = 3 is allowed. Hence, taking nf = 3, we get:
What is the energy in joules, required to shift the electron of the hydrogen atom from the first Bohr orbit to the fifth Bohr orbit and what is the wavelength of the light emitted when the electron returns to the ground state? The ground state electron energy is -2.18 x 10-11ergs.
Energy (E) of the nthBohr orbit of an atom is given by,
Where,
Z = atomic number of the atom
Ground state energy = โ 2.18 ื 10โ11ergs
= โ2.18 ื 10โ11ื 10โ7J
= โ 2.18 ื 10โ18J
Energy required to shift the electron from n= 1 to n = 5 is given as: ΔE = E5โ E1
The electron energy in hydrogen atom is given by En= (-2.18 x 10-18)/n2J. Calculate the energy required to remove an electron completely from the n= 2 orbit. What is the longest wavelength of light in cm that can be used to cause this transition?
Given,
Energy required for ionization from n= 2 is given by,
= 0.545 ื 10โ18J ΔE = 5.45 ื 10โ19J
Here, รยปis the longest wavelength causing the transition.
= 3647 ื 10โ10m
Calculate the wavelength of an electron moving with a velocity of 2.05 x 107 ms-1.
According to de Broglie's equation,
Where,
รยป = wavelength of moving particle
m = mass of particle v = velocity of particle h = Planck's constant
Substituting the values in the expression of รยป:
The mass of an electron is 9.1 x 10-31 kg. If its K.E. is 3.0 x 10-25 J, calculate its wavelength.
From de Broglie's equation,
Given,
Kinetic energy (K.E) of the electron = 3.0 ื 10โ25 J
Substituting the value in the expression of รยป:
Which of the following are isoelectronic species i.e., those having the same number of electrons? Na+, K+, Mg2+, Ca2+, S2-, Ar
Isoelectronic species have the same number of electrons. Number of electrons in sodium (Na) = 11
Number of electrons in (Na+) = 10
A positive charge denotes the loss of an electron. Similarly,
Number of electrons in K+ = 18 Number of electrons in Mg2+ = 10 Number of electrons in Ca2+ = 18
A negative charge denotes the gain of an electron by a species. Number of electrons in sulphur (S) = 16
∴ Number of electrons in S2- = 18
Number of electrons in argon (Ar) = 18 Hence, the following are isoelectronic species:
1) Na+ and Mg2+ (10 electrons each)
2)
(i) Write the electronic configurations of the following ions: (a) H- (b) Na+ (c) O2-(d) F-
(iii) Which atoms are indicated by the following configurations? (a) [He] 2s1 (b) [Ne] 3s2 3p3 (c) [Ar] 4s2 3d1.
(i) (a) H-ion
The electronic configuration of H atom is 1s1.
A negative charge on the species indicates the gain of an electron by it.
∴ Electronic configuration of H- = 1s2
(b) Na+ ion
The electronic configuration of Na atom is 1s2 2s2 2p6 3s1.
A positive charge on the species indicates the loss of an electron by it.
∴ Electronic configuration of Na+ = 1s2 2s2 2p6 3s0 or 1s2 2s2 2p6
(c) O2-ion
The electronic configuration of 0 atom is 1s2 2s2 2p4.
A dinegative charge on the species indicates that two electrons are gained by it.
∴ Electronic configuration of O2- ion = 1s2 2s2 p6
The electronic configuration of F atom is 1s2 2s2 2p5.
A negative charge on the species indicates the gain of an electron by it.
∴ Electron configuration of F- ion = 1s2 2s2 2p6
(ii) (a) 3s1
Completing the electron configuration of the element as 1s2 2s2 2p6 3s1.
∴ Number of electrons present in the atom of the element
= 2 + 2 + 6 + 1 = 11
∴ Atomic number of the element = 11
(b) 2p3
Completing the electron configuration of the element as
1s2 2s2 2p3.
∴ Number of electrons present in the atom of the element = 2 + 2 + 3 = 7
∴ Atomic number of the element = 7
(c) 3p5
Completing the electron configuration of the element as 1s2 2s2 2p5.
∴ Number of electrons present in the atom of the element = 2 + 2 + 5 = 9
∴ Atomic number of the element = 9
The electronic configuration of the element is [He] 2s1 = 1s2 2s1.
∴ Atomic number of the element = 3
Hence, the element with the electronic configuration [He] 2s1 is lithium (Li).
(b) [Ne] 3s2 3p3
The electronic configuration of the element is [Ne] 3s2 3p3= 1s2 2s2 2p6 3s2 3p3.
∴ Atomic number of the element = 15
Hence, the element with the electronic configuration [Ne] 3s2 3p3 is phosphorus (P).
(c) [Ar] 4s2 3d1
The electronic configuration of the element is [Ar] 4s2 3d1= 1s2 2s2 2p6 3s2 3p6 4s2 3d1.
∴ Atomic number of the element = 21
What is the lowest value of n that allows g orbitals to exist?
For g-orbitals, l = 4.
As for any value 'n' of principal quantum number, the Azimuthal quantum number (l) can have a value from zero to (n - 1).
An electron is in one of the 3d orbitals. Give the possible values of n, l and ml for this electron.
For the 3d orbital:
Principal quantum number (n) = 3 Azimuthal quantum number (l) = 2
An atom of an element contains 29 electrons and 35 neutrons. Deduce (i) the number of protons and (ii) the electronic configuration of the element.
(i) For an atom to be neutral, the number of protons is equal to the number of electrons.
∴ Number of protons in the atom of the given element = 29
(ii) The electronic configuration of the atom is 1s2 2s2 2p6 3s2 3p6 4s2 3d10.
Give the number of electrons in the species , H2 and
:
Number of electrons present in hydrogen molecule (H2) = 1 + 1 = 2
∴ Number of electrons in = 2 โ 1 = 1 H2:
Number of electrons in H2 = 1 + 1 = 2
:
Number of electrons present in oxygen molecule (O2) = 8 + 8 = 16
(i) An atomic orbital has n = 3. What are the possible values of l and ml ?
(iii) Which of the following orbitals are possible? 1p, 2s, 2p and 3f
(i) n = 3 (Given)
For a given value of n, l can have values from 0 to (n - 1).
∴ For n = 3
l = 0, 1, 2
For a given value of l, ml can have (2l + 1) values.
For l = 0, m = 0
l = 1, m = - 1, 0, 1
l = 2, m = - 2, - 1, 0, 1, 2
∴ For n = 3
l = 0, 1, 2
m0 = 0
m1 = - 1, 0, 1
m2 = - 2, - 1, 0, 1, 2
(ii) For 3d orbital, l = 2.
For a given value of l, mlcan have (2l + 1) values i.e., 5 values.
∴ For l = 2
m2 = - 2, - 1, 0, 1, 2
(iii) Among the given orbitals only 2s and 2p are possible. 1p and 3f cannot exist. For p-orbital, l = 1.
For a given value of n, l can have values from zero to (n - 1).
∴ For l is equal to 1, the minimum value of n is 2. Similarly,
For f-orbital, l = 4.
For l = 4, the minimum value of n is 5. Hence, 1p and 3f do not exist.
Using s, p, d notations, describe the orbital with the following quantum numbers.
(a) n = 1, l = 0; (b) n = 3; l =1 (c) n = 4; l = 2; (d) n = 4; l =3.
(a) n = 1, l = 0 (Given) The orbital is 1s.
(b) For n = 3 and l = 1 The orbital is 3p.
(c) For n = 4 and l = 2 The orbital is 4d.
(d) For n = 4 and l = 3 The orbital is 4f.
Explain, giving reasons, which of the following sets of quantum numbers are not possible.
a |
n = 0 |
l = 0 |
ml = 0 |
b |
n = 1 |
l = 0 |
ml = 0 |
c |
n = 1 |
l = 1 |
ml = 0 |
d |
n = 2 |
l = 1 |
ml = 0 |
(a) The given set of quantum numbers is not possible because the value of the principal quantum number (n) cannot be zero.
(b) The given set of quantum numbers is possible.
(c) The given set of quantum numbers is not possible.
For a given value of n, 'l' can have values from zero to (n - 1). For n = 1, l = 0 and not 1.
(d) The given set of quantum numbers is possible.
(e) The given set of quantum numbers is not possible. For n = 3,
l = 0 to (3 - 1)
l = 0 to 2 i.e., 0, 1, 2
(f)
How many electrons in an atom may have the following quantum numbers?(a)n = 4, (b) n = 3, l = 0
(a) Total number of electrons in an atom for a value of n = 2n2
∴ For n = 4,
Total number of electrons = 2 (4)2
= 32
The given element has a fully filled orbital as 1s2 2s2 2p6 3s2 3p6 4s2 3d10.
Hence, all the electrons are paired.
∴ Number of electrons (having n = 4 and ) = 16
(b)
Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around the orbit.
Where,
n = 1, 2, 3,
According to de Broglie's equation:
Substituting the value of 'mv' from expression (2) in expression (1):
What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n = 2 of He+ spectrum?
For He+ ion, the wave number associated with the Balmer transition, n = 4 to n = 2 is given by:
Where,
n1 = 2
n2 = 4
Z = atomic number of helium
According to the question, the desired transition for hydrogen will have the same wavelength as that of He+.
By hit and trail method, the equality given by equation (1) is true only when
n1 = 1and n2 = 2.
Calculate the energy required for the process
Answer :
Energy associated with hydrogen-like species is given by,
For ground state of hydrogen atom,
For the given process,
An electron is removed from n = 1 to n = ∞.
∴ The energy required for the process
If the diameter of a carbon atom is 0.15 nm, calculate the number of carbon atoms which can be placed side by side in a straight line across length of scale of length 20 cm long.
1 m = 100 cm 1 cm = 10โ2 m
Length of the scale = 20 cm
= 20 ื 10โ2 m
Diameter of a carbon atom = 0.15 nm
= 0.15 ื 10โ9 m
One carbon atom occupies 0.15 ื 10โ9 m.
∴ Number of carbon atoms that can be placed in a straight line
2 x 108 atoms of carbon are arranged side by side. Calculate the radius of carbon atom if the length of this arrangement is 2.4 cm.
Length of the given arrangement = 2.4 cm Number of carbon atoms present = 2 ื 108
∴ Diameter of carbon atom
The diameter of zinc atom is .Calculate (a) radius of zinc atom in pm and (b) number of atoms present in a length of 1.6 cm if the zinc atoms are arranged side by side lengthwise.
(a) Radius of zinc atom
(b) Length of the arrangement = 1.6 cm
= 1.6 ื 10โ2 m
Diameter of zinc atom = 2.6 ื 10โ10 m
∴ Number of zinc atoms present in the arrangement
A certain particle carries 2.5 x 10-16C of static electric charge. Calculate the number of electrons present in it.
Charge on one electron = 1.6022 ื 10โ19 C
⇒ 1.6022 ื 10โ19C charge is carried by 1 electron.
In Milikan's experiment, static electric charge on the oil drops has been obtained by shining X-rays. If the static electric charge on the oil drop is -1.282 x 10-18C, calculate the number of electrons present on it.
Charge on the oil drop = 1.282 ื10โ18C Charge on one electron = 1.6022 ื 10โ19C
In Rutherford's experiment, generally the thin foil of heavy atoms, like gold, platinum etc. have been used to be bombarded by the α-particles. If the thin foil of light atoms like aluminium etc. is used, what difference would be observed from the above results?
A thin foil of lighter atoms will not give the same results as given with the foil of heavier atoms.
Symbols can be written, whereas symbols are not acceptable. Answer briefly.
The general convention of representing an element along with its atomic mass (A) and atomic number (Z) is Hence, is acceptable but is not acceptable.
An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the atomic symbol.
Let the number of protons in the element be x.
∴ Number of neutrons in the element
= x + 31.7% of x
= x + 0.317 x
= 1.317 x
According to the question,
Mass number of the element = 81
Hence, the number of protons in the element i.e., x is 35.
Since the atomic number of an atom is defined as the number of protons present in its nucleus, the atomic number of the given element is 35.
An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the atomic symbol.
Let the number of protons in the element be x.
∴ Number of neutrons in the element
= x + 31.7% of x
= x + 0.317 x
= 1.317 x
According to the question,
Mass number of the element = 81
Hence, the number of protons in the element i.e., x is 35.
Since the atomic number of an atom is defined as the number of protons present in its nucleus, the atomic number of the given element is 35.
An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than electrons. Assign the symbol to this ion.
Let the number of electrons present in ion
∴ Number of neutrons in it = x + 30.4% of x = 1.304 x
Since the ion is tripositive,
⇒ Number of electrons in neutral atom = x + 3
∴ Number of protons in neutral atom = x + 3 Given,
∴ Number of protons = x + 3 = 23 + 3 = 26
Arrange the following type of radiations in increasing order of frequency: (a) radiation from microwave oven
Answer :
The increasing order of frequency is as follows:
Radiation from FM radio < amber light < radiation from microwave oven < X- rays < cosmic rays The increasing order of wavelength is as follows:
Nitrogen laser produces a radiation at a wavelength of 337.1 nm. If the number of photons emitted is 5.6 x 1024, calculate the power of this laser.
Power of laser = Energy with which it emits photons
Power Where,
N = number of photons emitted h = Planck's constant
c = velocity of radiation
รยป = wavelength of radiation
Substituting the values in the given expression of Energy (E):
E
= 0.3302 ื 107 J
= 3.33 ื 106 J
Neon gas is generally used in the sign boards. If it emits strongly at 616 nm, calculate (a) the frequency of emission, (b) distance traveled by this radiation in 30 s (c) energy of quantum and (d) number of quanta present if it produces 2 J of energy.
Wavelength of radiation emitted = 616 nm = 616 ื 10โ9 m (Given)
(a)
Where,
c = velocity of radiation
รยป = wavelength of radiation
= 4.87 ื 108 ื 109 ื 10โ3 sโ1
รยฝ = 4.87 ื 1014 sโ1
Frequency of emission (รยฝ) = 4.87 ื 1014 sโ1
(b) Velocity of radiation, (c) = 3.0 ื 108 msโ1 Distance travelled by this radiation in 30 s
= (3.0 ื 108 msโ1) (30 s)
= 9.0 ื 109 m
(c) Energy of quantum (E) = hรยฝ (6.626 ื 10โ34 Js) (4.87 ื 1014 sโ1)
Energy of quantum (E) = 32.27 ื 10โ20 J
(d) Energy of one photon (quantum) = 32.27 ื 10โ20 J
Therefore, 32.27 ื 10โ20 J of energy is present in 1 quantum. Number of quanta in 2 J of energy
= 6.19 ื1018
In astronomical observations, signals observed from the distant stars are generally weak. If the photon detector receives a total of 3.15 x 10-18J from the radiations of 600 nm, calculate the number of photons received by the detector.
Where,
รยป = wavelength of radiation
h = Planck's constant
c = velocity of radiation
Substituting the values in the given expression of E:
E E = 3.313 ื 10โ19 J
Energy of one photon = 3.313 ื 10โ19 J
= 9.5
Lifetimes of the molecules in the excited states are often measured by using pulsed radiation source of duration nearly in the nano second range. If the radiation source has the duration of 2 ns and the number of photons emitted during the pulse source is 2.5 x 1015, calculate the energy of the source.
รยฝ = 5.0 ื 108 sโ1
Energy (E) of source = Nhรยฝ
Where,
N = number of photons emitted h = Planck's constant
รยฝ = frequency of radiation
Substituting the values in the given expression of (E):
E = (2.5 ื 1015) (6.626 ื 10โ34 Js) (5.0 ื 108 sโ1)
E = 8.282 ื 10โ10 J
The longest wavelength doublet absorption transition is observed at 589 and 589.6 nm. Calculate the frequency of each transition and energy difference between two excited states.
For รยป1 = 589 nm
Frequency of transition (รยฝ1) = 5.093 ื 1014 sโ1 Similarly, for รยป2 = 589.6 nm
Frequency of transition
Frequency of transition (รยฝ2) = 5.088 ื 1014 sโ1 Energy difference (ΔE) between excited states = E1 โ E2 Where,
E2 = energy associated with รยป2 E1 = energy associated with รยป1 ΔE= hรยฝ1โ hรยฝ2
= h(รยฝ1โ รยฝ2)
= (6.626 ื 10โ34 Js) (5.093 ื 1014 โ 5.088 ื 1014)sโ1
= (6.626 ื 10โ34 J) (5.0 ื 10โ3ื 1014) ΔE = 3.31 ื 10โ22J
The work function for caesium atom is 1.9 eV. Calculate (a) the threshold wavelength and (b) the threshold frequency of the radiation. If the caesium element is irradiated with a wavelength 500 nm, calculate the kinetic energy and the velocity of the ejected photoelectron.
It is given that the work function (W0) for caesium atom is 1.9 eV.
(a)
Where,
รยป0 = threshold wavelength h = Planck's constant
c = velocity of radiation
6.53 ื 10โ7 m
Hence, the threshold wavelength is 653 nm.
(b)
Where,
รยฝ0= threshold frequency h = Planck's constant
(1 eV = 1.602 ื 10โ19J)
รยฝ0= 4.593 ื 1014sโ1
Hence, the threshold frequency of radiation (รยฝ0) is 4.593 ื 1014sโ1.
(c) According to the question:
Wavelength used in irradiation (รยป) = 500 nm Kinetic energy = h (รยฝ โ รยฝ0)
= 9.3149 ื 10โ20 J
Kinetic energy of the ejected photoelectron = 9.3149 ื 10โ20J Since K.E
v = 4.52 ื 105 msโ1
Following results are observed when sodium metal is irradiated with different wavelengths. Calculate (a) threshold wavelength and, (b) Planck's constant.
รยป (nm) |
500 |
450 |
400 |
v ื 10โ5 (cm sโ1) |
2.55 |
4.35 |
5.35 |
Three different equalities can be formed by the given value as:
Threshold wavelength = 540 nm
The ejection of the photoelectron from the silver metal in the photoelectric effect experiment can be stopped by applying the voltage of 0.35 V when the radiation 256.7 nm is used. Calculate the work function for silver metal.
From the principle of conservation of energy, the energy of an incident photon (E) is equal to the sum of the work function (W0) of radiation and its kinetic energy (K.E) i.e.,
E = W0 + K.E
⇒ W0= E โ K.E
Energy of incident photon (E) Where,
c= velocity of radiation h= Planck's constant
รยป = wavelength of radiation
E = 4.83 eV
The potential applied to silver metal changes to kinetic energy (K.E) of the photoelectron. Hence,
K.E = 0.35 V
K.E = 0.35 eV
Work function, W0= E โ K.E
= 4.83 eV โ 0.35 eV
If the photon of the wavelength 150 pm strikes an atom and one of its inner bound electrons is ejected out with a velocity of 1.5 x 107ms-1, calculate the energy with which it is bound to the nucleus.
= 10.2480 ื 10โ17 J
= 1.025 ื 10โ16J
Hence, the energy with which the electron is bound to the nucleus can be obtained as:
= E โ K.E
= 13.252 ื 10โ16J โ 1.025 ื 10โ16J
Emission transitions in the Paschen series end at orbit n= 3 and start from orbit n and can be represented as v = 3.29 x 1015(Hz) [1/32- 1/n2]
Answer :
Wavelength of transition = 1285 nm
= 1285 ื 10โ9m (Given)
(Given)
Since
รยฝ = 2.33 ื 1014 sโ1
Substituting the value of รยฝ in the given expression,
n= 4.98
n รขโฐห5
Hence, for the transition to be observed at 1285 nm, n= 5. The spectrum lies in the infra-red region.
Calculate the wavelength for the emission transition if it starts from the orbit having radius 1.3225 nm and ends at 211.6 pm. Name the series to which this transition belongs and the region of the spectrum.
The radius of the nth orbit of hydrogen-like particles is given by,
For radius (r1) = 1.3225 nm
= 1.32225 ื 10โ9 m
= 1322.25 ื 10โ12 m
= 1322.25 pm
Similarly,
⇒ n1 = 5 and n2 = 2
Thus, the transition is from the 5th orbit to the 2nd orbit. It belongs to the Balmer series. Wave number for the transition is given by,
1.097 ื 107 mโ1
= 2.303 ื 106 mโ1
Wavelength (รยป) associated with the emission transition is given by,
= 0.434 ื10โ6 m
รยป = 434 nm
โThis transition
belongs to Balmer series and comes in the visible region of the spectrum.
Dual behaviour of matter proposed by de Broglie led to the discovery of electron microscope often used for the highly magnified images of biological molecules and other type of material. If the velocity of the electron in this microscope is 1.6 x 106ms-1, calculate de Broglie wavelength associated with this electron.
From de Broglie's equation,
= 4.55 ื 10โ10m รยป = 455 pm
de Broglie's
wavelength associated with the electron is 455 pm.
Similar to electron diffraction, neutron diffraction microscope is also used for the determination of the structure of molecules. If the wavelength used here is 800 pm, calculate the characteristic velocity associated with the neutron.
From de Broglie's equation,
Where,
v = velocity of particle (neutron)
h = Planck's constant
m = mass of particle (neutron) รยป = wavelength
Substituting the values in the expression of velocity (v),
= 4.94 ื 102msโ1
v= 494 msโ1
Velocity associated with the neutron = 494 msโ1
If the velocity of the electron in Bohr's first orbit is 2.19 x 106ms-1, calculate the de Broglie wavelength associated with it.
According to de Broglie's equation,
Where,
รยป = wavelength associated with the electron
h= Planck's constant m= mass of electron v= velocity of electron
Substituting the values in the expression of รยป:
รยป = 332 pm
Wavelength associated with the electron
= 332 pm
The velocity associated with a proton moving in a potential difference of 1000 V is 4.37 x 105ms-1. If the hockey ball of mass 0.1 kg is moving with this velocity, calculate the wavelength associated with this velocity.
According to de Broglie's expression,
Substituting the values in the expression,
If the position of the electron is measured within an accuracy of + 0.002 nm, calculate the uncertainty in the momentum of the electron. Suppose the momentum of the electron is h/4รโฌm x 0.05 nm, is there any problem in defining this value.
From Heisenberg's uncertainty principle,
Where,
Δx = uncertainty in position of the electron
Δp = uncertainty in momentum of the electron Substituting the values in the expression of Δp:
= 2.637 ื 10โ23 Jsmโ1
Δp = 2.637 ื 10โ23 kgmsโ1 (1 J = 1 kgms2sโ1)
Uncertainty in the momentum of the electron = 2.637 ื 10โ23 kgmsโ1.
Actual momentum
= 1.055 ื 10โ24 kgmsโ1
Since the
magnitude of the actual momentum
is smaller than the uncertainty, the value cannot be
defined.
The quantum numbers of six electrons are given below. Arrange them in order of increasing energies. If any of these combination(s) has/have the same energy lists:
1. n= 4, l = 2, ml= -2 , ms= -1/2
2. n= 3, l = 2, ml= 1 , ms= +1/2
3. n= 4, l = 1, ml= 0 , ms= +1/2
4. n = 3, l= 2, ml= -2 , ms= -1/2
5. n = 3, l= 1, ml= -1 , ms= +1/2
6. n = 4, l= 1, ml= 0 , ms= +1/2
For n = 4 and l = 2, the orbital occupied is 4d. For n = 3 and l = 2, the orbital occupied is 3d. For n = 4 and l = 1, the orbital occupied is 4p.
Hence, the six electrons i.e., 1, 2, 3, 4, 5, and 6 are present in the 4d, 3d, 4p, 3d, 3p, and 4p orbitals respectively. Therefore, the increasing order of energies is 5(3p) < 2(3d) = 4(3d) < 3(4p) = 6(4p) < 1 (4d).
The bromine atom possesses 35 electrons. It contains 6 electrons in 2porbital, 6 electrons in 3porbital and 5 electrons in 4porbital. Which of these electron experiences the lowest effective nuclear charge?
Nuclear charge experienced by an electron (present in a multi-electron atom) is dependant upon the distance between the nucleus and the orbital, in which the electron is present. As the distance increases, the effective nuclear charge also decreases.
Among p-orbitals, 4p orbitals are farthest from the nucleus of bromine atom
with (+35) charge. Hence, the electrons in the 4p orbital will experience the lowest effective
nuclear charge. These electrons are shielded by electrons present
in the 2pand 3p orbitals along
with the s-orbitals. Therefore, they will experience the lowest nuclear
charge.
Among the following pairs of orbitals which orbital will experience the larger effective nuclear charge? (i) 2s and 3s, (ii) 4d and 4f, (iii) 3d and 3p
Nuclear charge is defined as the net positive charge experienced by an electron in the orbital of a multi-electron atom. The closer the orbital, the greater is the nuclear charge experienced by the electron (s) in it.
(i) The electron(s) present in the 2s orbital will experience greater nuclear charge (being closer to the nucleus) than the electron(s) in the 3s orbital.
(ii) 4d will experience greater nuclear charge than 4f since 4d is closer to the nucleus.
(iii)
3p will experience greater
nuclear charge since it is closer
to the nucleus than 3f.
The unpaired electrons in Al and Si are present in 3p orbital. Which electrons will experience more effective nuclear charge from the nucleus?
Nuclear charge is defined as the net positive charge experienced by an electron in a multi-electron atom.
The higher the atomic number, the higher is the
nuclear charge. Silicon has 14 protons while aluminium has 13 protons. Hence, silicon has a larger
nuclear charge of (+14) than aluminium, which has a nuclear charge of (+13). Thus,
the electrons in the 3p orbital of silicon
will experience a more
effective nuclear charge than aluminium.
Indicate the number of unpaired electrons in: (a) P, (b) Si, (c) Cr, (d) Fe and (e) Kr.
(a) Phosphorus (P):
Atomic number = 15
The electronic configuration of P is:
1s2 2s2 2p6 3s2 3p3
From the orbital picture, phosphorus has three unpaired electrons.
(b) Silicon (Si):
Atomic number = 14
The electronic configuration of Si is:
1s2 2s2 2p6 3s2 3p2
From the orbital picture, silicon has two unpaired electrons.
(c) Chromium (Cr):
Atomic number = 24
The electronic configuration of Cr is:
1s2 2s2 2p6 3s2 3p6 4s1 3d5
From the orbital picture, chromium has six unpaired electrons.
(d) Iron (Fe):
Atomic number = 26
The electronic configuration is:
1s2 2s2 2p6 3s2 3p6 4s2 3d6
From the orbital picture, iron has four unpaired electrons.
(e) Krypton (Kr):
Atomic number = 36
The electronic configuration is:
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 The orbital picture of krypton is:
Since all orbitals
are fully occupied, there are no unpaired electrons in krypton.
(a) How many sub-shells are associated with n = 4? (b) How many electrons will be present in the sub-shells having ms value of -1/2 for n = 4?
(a) n = 4 (Given)
For a given value of 'n', 'l' can have values from zero to (n โ 1).
∴ l = 0, 1, 2, 3
Thus, four sub-shells are associated with n = 4, which are s, p, d and f.
(b) Number of orbitals in the nth shell = n2
For n = 4
Number of orbitals = 16
If each orbital is taken fully, then it will have 1 electron with msvalue of .
∴ Number of electrons with ms value of is 16.